\newproblem{lay:2_4_16}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.4.16}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A=\begin{pmatrix}A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}$. If $A_{11}$ is invertible, then
	the matrix $S=A_{22}-A_{21}A_{11}^{-1}A_{12}$ is called the Schur complement of $A_{11}$. Likewise, if $A_{22}$
	is invertible, the matrix $A_{11}-A_{12}A_{22}^{-1}A_{21}$ is called the Schur complement of $A_{22}$. Suppose
	$A_{11}$ is invertible. Find $X$ and $Y$ such that
	\begin{center}
		$\begin{pmatrix}A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}=
		 \begin{pmatrix}I & 0 \\ X & I \end{pmatrix}\begin{pmatrix}A_{11} & 0 \\ 0 & S \end{pmatrix}
		 \begin{pmatrix}I & Y \\ 0 & I \end{pmatrix}$
	\end{center}
}{
  % Solution
	Let us multiply the matrices on the right
	\begin{center}
		$\begin{pmatrix}I & 0 \\ X & I \end{pmatrix}\begin{pmatrix}A_{11} & 0 \\ 0 & S \end{pmatrix}
		 \begin{pmatrix}I & Y \\ 0 & I \end{pmatrix} =
		 \begin{pmatrix}A_{11} & 0 \\ XA_{11} & S \end{pmatrix}\begin{pmatrix}I & Y \\ 0 & I \end{pmatrix}=
		 \begin{pmatrix}A_{11} & A_{11}Y \\ XA_{11} & XA_{11}Y+S \end{pmatrix}$
	\end{center}
	Comparing this product to $A$ we derive the following equations:
	\begin{center}
		$A_{11}Y=A_{12}$ \\
		$XA_{11}=A_{21}$ \\
		$XA_{11}Y+S=A_{22}$ \\
	\end{center}
	That are solved like
	\begin{center}
		$Y=A_{11}^{-1}A_{12}$ \\
		$X=A_{21}A_{11}^{-1}$ \\
	\end{center}
	We need to check that the last equation is verified
	\begin{center}
		$XA_{11}Y+S=A_{22}$ \\
		$(A_{21}A_{11}^{-1})A_{11}(A_{11}^{-1}A_{12})+(A_{22}-A_{21}A_{11}^{-1}A_{12})=A_{22}$ \\
		$A_{21}A_{11}^{-1}A_{12}+A_{22}-A_{21}A_{11}^{-1}A_{12}=A_{22}$ \\
		$A_{22}=A_{22}$ \\
	\end{center}
}
\useproblem{lay:2_4_16}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
